A certain hydrocarbon on complete combustion at s.t.p produced 89.6dm\(^3\) of CO\(_2\) and 54g of water. The hydrocarbon should be?
C\(_6\)H\(_6\)
C\(_4\)H\(_{10}\)
C\(_5\)H\(_{10}\)
C\(_4\)H\(_6\)
Explanation
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In the question above an Hydrocarbon combust to give CO2 and H20
let Hydrocarbon be
CxHy + x+Y/4O2= xCO2 + Y/2H2O
mass of C0=44g and H2O=18g
at STP vol= 22.4
therefore, 1mole of CO2 contains 44g and 22.4dm³ at STP
1mole=22.4dm³
xmole=89.6dm³
cross multiplying x=89.6/22.4 =4mole of CO2 produce
1mole of H2O= 18g
Xmole=56g
cross multiplying
X=56/18 = 3mole of H20
then....
CxHy +X+y/4O2= 4CO2+ 3H2O
balancing
C4H6 + 6O2 = 4CO2 + 3H2O
what if I do it this way
89.6cmd³ of CO2
at s t p =22.4
89.6/22.4=4
56g of H20
1x2+16=18
56/18=3
hydrogen is 2 so 2x3=6
C4H6💪💯
Shortcut method🤘
For CO2
n=V/22.4
n=89.6/22.4
n=4
To find our "C", multiply 4 above☝️ by the number of carbon atoms in CO2. The number of carbon atoms in CO2 is 1
So, 4×1 =4
Therefore, C=4
~~~~~~~~~~~~~~~~~~~~~
For H20
n=mass/molar mass
n=54/18
n=3
To find our "H", multiply 3 above ☝️ by number of hydrogen atoms in H2O. The number of hydrogen atoms in H2O is 2.
So, 3×2 =6
Therefore, H= 6
Finally, C4H6
Try solve this one,
A certain hydrocarbon on complete combustion at stp produced 44.8dm³ of CO2 and 36g of H2O. The hydrocarbon should be?....
A. C2H4
B. C3H8
C. C2H6
D. C3H6
Goodluck
That formula is the general formula for the combustion of alkanes, check it out on google or under alkanes in organic chemistry section of your chemistry textbook
